If the distance between the points `(2,a,1)and(-1,-1,1)` is 5, then the value(s) of a is

Options

A. `3, 5`
B. `-3, 5`
C. `3, -5`
D. `-3, -5`

Answer

To solve the problem, we need to find the value(s) of \( a \) such that the distance between the points \( (2, a, 1) \) and \( (-1, -1, 1) \) is equal to 5.

### Step-by-step Solution:

1. **Identify the Points**:
The two points are \( P(2, a, 1) \) and \( Q(-1, -1, 1) \).

2. **Distance Formula**:
The distance \( d \) between two points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) in 3D space is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

3. **Substituting the Points**:
Substitute the coordinates of points \( P \) and \( Q \) into the distance formula:
\[ d = \sqrt{((-1) - 2)^2 + ((-1) - a)^2 + (1 - 1)^2} \]
This simplifies to:
\[ d = \sqrt{(-3)^2 + (-1 - a)^2 + 0^2} \]
\[ d = \sqrt{9 + (a + 1)^2} \]

4. **Setting the Distance Equal to 5**:
We know the distance is equal to 5, so we set up the equation:
\[ \sqrt{9 + (a + 1)^2} = 5 \]

5. **Squaring Both Sides**:
To eliminate the square root, square both sides:
\[ 9 + (a + 1)^2 = 25 \]

6. **Isolating the Quadratic Term**:
Subtract 9 from both sides:
\[ (a + 1)^2 = 16 \]

7. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ a + 1 = 4 \quad \text{or} \quad a + 1 = -4 \]

8. **Solving for \( a \)**:
For the first equation:
\[ a + 1 = 4 \implies a = 3 \]
For the second equation:
\[ a + 1 = -4 \implies a = -5 \]

9. **Final Values**:
Therefore, the values of \( a \) are:
\[ a = 3 \quad \text{and} \quad a = -5 \]

### Summary of the Solution:
The values of \( a \) that satisfy the condition are \( a = 3 \) and \( a = -5 \).